Question

If $x={\tan }^{-1}1-{\cos }^{-1}[-\frac{1}{2}]+{\sin }^{-1}\frac{1}{2};y=\cos \left[\frac{1}{2}{\cos }^{-1}\left[\frac{1}{8}\right]\right]$ then -

• A. $x=\pi y$
• B. $y=\pi x$
• C. $\tan x=-\left(4/3\right)y$
• D. $\tan x=\left(4/3\right)y$

Answer 1

Answer: (C)

$\tan x=-\left(4/3\right)y$

Given, $x={\tan }^{-1}1-{\cos }^{-1}[-\frac{1}{2}]+{\sin }^{-1}\frac{1}{2}$ and $y=\cos \left[\frac{1}{2}{\cos }^{-1}\left[\frac{1}{8}\right]\right]$

here, $x={\tan }^{-1}1-{\cos }^{-1}[-\frac{1}{2}]+{\sin }^{-1}\frac{1}{2}$

$x=\frac{\pi }{4}-\frac{2\pi }{3}+\frac{\pi }{6}$

$\Rightarrow x=\frac{-3\pi }{12}=\frac{-\pi }{4}$
$y=\cos \left[\frac{1}{2}{\cos }^{-1}\left[\frac{1}{8}\right]\right]$

Let, ${\cos }^{-1}\left[\frac{1}{8}\right]$$=\theta$

$\Rightarrow \cos \theta =\frac{1}{8}$

$\Rightarrow \cos \frac{\theta }{2}=\sqrt{\frac{1+\cos \theta }{2}}=\frac{3}{4}=y$

$\Rightarrow \tan x=\tan \frac{-\pi }{4}=1=\frac{-4}{3}\times \frac{3}{4}$ as in option c.